[EMAIL PROTECTED] wrote:
> Hi everyone.
>
> I'm trying to work with very simple data structures but I'm stuck in the very
> first steps. If someone has the luxury of a few minutes and can give an
> advice how to resolve this, I'll really appreciate it.
>
> 1- I have a list of tuples like this:
> lista= [(162, 141, 3), (162, 141, 3), (162, 141, 3), (168, 141, 2), (168,
> 141, 2), (168, 141, 2), (201, 141, 1), (213, 141, 1), (203, 141, 1), (562,
> 142, 4), (562, 142, 4), (562, 142, 4), (568, 142, 2), (568, 142, 2), (568,
> 142, 2), (501, 142, 1), (513, 142, 1), (503, 142, 1)]
> and I want to end with a dict like this:
> {141: {1: [203, 213, 201], 2: [168, ], 3: [162, ]}, 142: {1: [503, 513, 501],
> 2: [568, ], 4: [562, ]}}
> the logic of the final output:
> a) the outer dict's key is a set() of the 2rd value of the input.
> b) the inner dict's key is a set() of the 3th value for tuples which 3rd
> value equals a).
> c) the inner list will be fill up with the 1st value of every tuple which
> 3rd value equals b) and its 2rd value equals a).
>
> So far, the only thing it seems I can achieve is the first part:
> outer_dict = dict([(x,dict()) for x in set(row[1] for row in lista)])
>
>>From then on, I'm starting to get tired after several successful failures (I
>>tried with itertools, with straight loops ...) and I don't know which can be
>>the easier way to get that final output.
>
> Thanks in advance.
>
>
d={}
for a, b, c in lista:
if d.has_key(b):
if d[b].has_key(c):
if a not in d[b][c]: d[b][c].append(a)
else:
d[b][c]=[a]
else:
d[b]={c:[a]}
print d
-Larry Bates
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